Question

What is the definite integral of log (tan x) with range from 90 to 0

Answer 1

Substituting x with π/2 - x gives the equivalent integral,

`\displaystyle\int_0^(\frac\pi2)\log(\tan(x))\,\mathrm dx=-\int_(\frac\pi2)^0\log(\cot(x))(-\mathrm dx)=\int_0^(\frac\pi2)\log(\cot(x))\,\mathrm dx`

So if we let J denote the value of the integral, we have

`J=\displaystyle\int_0^(\frac\pi2)\log(\tan (x))\,\mathrm dx`

`J=\displaystyle\int_0^(\frac\pi2)\log(\cot (x))\,\mathrm dx`

`\implies 2J=\displaystyle\int_0^(\frac\pi2)\left(\log(\tan (x))+\log(\cot (x))\right)\,\mathrm dx`

Condensing the logarithms, we have

log(tan(x)) + log(cot(x)) = log(tan(x) cot(x)) = log(1) = 0

since cot(x) = 1/tan(x), which means

`2J=\displaystyle\int_0^(\frac\pi2)0\,\mathrm dx=0`

and so the original integral has a value of J = 0.